NCERT Limits and derivatives

Chapter

13

he

LIMITS AND DERIVATIVES

bl

13.1 Introduction

is

With the Calculus as a key, Mathematics can be successfully applied to the explanation of the course of Nature – WHITEHEAD

© no N C tt E o R be T re pu

This chapter is an introduction to Calculus. Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. First, we give an intuitive idea of derivative (without actually defining it). Then we give a naive definition of limit and study some algebra of limits. Then we come back to a definition of derivative and study some algebra of derivatives. We also obtain derivatives of certain standard functions.

13.2 Intuitive Idea of Derivatives

Sir Issac Newton (1642-1727)

Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of 4.9t2 metres in t seconds, i.e., distance s in metres covered by the body as a function of time t in seconds is given by s = 4.9t2.

The adjoining Table 13.1 gives the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff.

The objective is to find the veloctiy of the body at time t = 2 seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at t = 2 seconds and hope that these throw some light on the velocity at t = 2 seconds.

Average velocity between t = t1 and t = t2 equals distance travelled between t = t1 and t = t2 seconds divided by (t2 – t1). Hence the average velocity in the first two seconds

MATHEMATICS

Table 13.1

Distance travelled between t2 = 2 and t1 = 0 = Time interval (t2 − t1 ) =

t 0 1 1.5 1.8 1.9 1.95 2 2.05 2.1 2.2 2.5 3 4

(19.6 − 0 ) m = 9.8 m / s . ( 2 − 0) s

Similarly, the average velocity between t = 1 and t = 2 is

bl

© no N C tt E o R be T re pu

Likewise we compute the average velocitiy between t = t1 and t = 2 for various t1. The following Table 13.2 gives the average velocity (v), t = t1 seconds and t = 2 seconds.

is

(19.6 – 4.9 ) m = 14.7 m/s ( 2 − 1) s

s 0 4.9 11.025 15.876 17.689 18.63225 19.6 20.59225 21.609 23.716 30.625 44.1 78.4

he

282

Table 13.2

t1

0

1

1.5

1.8

1.9

1.95

1.99

v

9.8

14.7

17.15

18.62

19.11

19.355

19.551

From Table 13.2, we observe that the average velocity is gradually increasing.

As we make the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2. Hoping that nothing really dramatic happens between 1.99 seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just above 19.551m/s.

This conclusion is somewhat strengthened by the following set of computation.

Compute the average velocities for various time intervals starting at t = 2 seconds. As before the average velocity v between t = 2 seconds and t = t2 seconds is =

Distance travelled between 2 seconds and t2 seconds t2 − 2

=

Distance travelled in t2 seconds − Distance travelled in 2 seconds t2 − 2

LIMITS AND DERIVATIVES

=

283

Distance travelled in t2 seconds − 19.6 t2 − 2

Table 13.3

4

v

29.4

3

2.5

2.2

2.1

2.05

2.01

24.5 22.05 20.58 20.09 19.845 19.649

is

t2

he

The following Table 13.3 gives the average velocity v in metres per second between t = 2 seconds and t2 seconds.

Š no N C tt E o R be T re pu

bl

Here again we note that if we take smaller time intervals starting at t = 2, we get better idea of the velocity at t = 2.

In the first set of computations, what we have done is to find average velocities in increasing time intervals ending at t = 2 and then hope that nothing dramatic happens just before t = 2. In the second set of computations, we have found the average velocities decreasing in time intervals ending at t = 2 and then hope that nothing dramatic happens just after t = 2. Purely on the physical grounds, both these sequences of average velocities must approach a common limit. We can safely conclude that the velocity of the body at t = 2 is between 19.551m/s and 19.649 m/s. Technically, we say that the instantaneous velocity at t = 2 is between 19.551 m/s and 19.649 m/s. As is well-known, velocity is the rate of change of displacement. Hence what we have accomplished is the following. From the given data of distance covered at various time instants we have estimated the rate of change of the distance at a given instant of time. We say that the derivative of the distance function s = 4.9t2 at t = 2 is between 19.551 and 19.649.

An alternate way of viewing this limiting process is shown in Fig 13.1.

This is a plot of distance s of the body from the top of the cliff versus the time t elapsed. In the limit as the sequence of time intervals h1, h2, ..., approaches zero, the sequence of average velocities approaches the same limit as does the Fig 13.1 sequence of ratios

284

MATHEMATICS

C1B1 C2 B2 C3 B3 , , , ...

AC1 AC2 AC3

he

where C1B1 = s1 – s0 is the distance travelled by the body in the time interval h1 = AC1, etc. From the Fig 13.1 it is safe to conclude that this latter sequence approaches the slope of the tangent to the curve at point A. In other words, the instantaneous velocity v(t) of a body at time t = 2 is equal to the slope of the tangent of the curve s = 4.9t2 at t = 2.

13.3 Limits

© no N C tt E o R be T re pu

lim f ( x ) = 0

bl

is

The above discussion clearly points towards the fact that we need to understand limiting process in greater clarity. We study a few illustrative examples to gain some familiarity with the concept of limits.

Consider the function f(x) = x2. Observe that as x takes values very close to 0, the value of f(x) also moves towards 0 (See Fig 2.10 Chapter 2). We say x →0

(to be read as limit of f (x) as x tends to zero equals zero). The limit of f (x) as x tends to zero is to be thought of as the value f (x) should assume at x = 0.

In general as x → a, f (x) → l, then l is called limit of the function f (x) which is

f ( x) = l . symbolically written as lim x→a

Consider the following function g(x) = |x|, x ≠ 0. Observe that g(0) is not defined.

Computing the value of g(x) for values of x very near to 0, we see that the value of g(x) moves towards 0. So, lim g(x) = 0. This is intuitively x→0 clear from the graph of y = |x| for x ≠ 0.

(See Fig 2.13, Chapter 2).

Consider the following function.

x2 − 4 , x≠2. x−2 Compute the value of h(x) for values of x very near to 2 (but not at 2). Convince yourself that all these values are near to 4. This is somewhat strengthened by considering the graph of the function y = h(x) given here (Fig 13.2). h( x) =

Fig 13.2

LIMITS AND DERIVATIVES

285

he

In all these illustrations the value which the function should assume at a given point x = a did not really depend on how is x tending to a. Note that there are essentially two ways x could approach a number a either from left or from right, i.e., all the values of x near a could be less than a or could be greater than a. This naturally leads to two limits – the right hand limit and the left hand limit. Right hand limit of a function f(x) is that value of f(x) which is dictated by the values of f(x) when x tends to a from the right. Similarly, the left hand limit. To illustrate this, consider the function

bl

Graph of this function is shown in the Fig 13.3. It is clear that the value of f at 0 dictated by values of f(x) with x ≤ 0 equals 1, i.e., the left hand limit of f (x) at 0 is

is

⎧1, x ≤ 0 f ( x) = ⎨ ⎩ 2, x > 0

lim f ( x) =1 . x→0

© no N C tt E o R be T re pu

Similarly, the value of f at 0 dictated by values of f (x) with x > 0 equals 2, i.e., the right hand limit of f (x) at 0 is lim f ( x) = 2 .

x → 0+

Fig 13.3

In this case the right and left hand limits are different, and hence we say that the limit of f (x) as x tends to zero does not exist (even though the function is defined at 0).

Summary

We say xlim f(x) is the expected value of f at x = a given the values of f near →a –

x to the left of a. This value is called the left hand limit of f at a.

f ( x ) is the expected value of f at x = a given the values of We say xlim →a+

f near x to the right of a. This value is called the right hand limit of f(x) at a.

If the right and left hand limits coincide, we call that common value as the limit of f(x) at x = a and denote it by lim f(x). x →a

Illustration 1 Consider the function f(x) = x + 10. We want to find the limit of this function at x = 5. Let us compute the value of the function f(x) for x very near to 5.

Some of the points near and to the left of 5 are 4.9, 4.95, 4.99, 4.995. . ., etc. Values of the function at these points are tabulated below. Similarly, the real number 5.001,

286

MATHEMATICS

5.01, 5.1 are also points near and to the right of 5. Values of the function at these points are also given in the Table 13.4.

Table 13.4

4.9

4.95

4.99

4.995

5.001

5.01

5.1

f(x)

14.9

14.95

14.99

14.995

15.001

15.01

15.1

he

x

bl

lim f ( x ) = 15 .

x →5 –

is

From the Table 13.4, we deduce that value of f(x) at x = 5 should be greater than 14.995 and less than 15.001 assuming nothing dramatic happens between x = 4.995 and 5.001. It is reasonable to assume that the value of the f(x) at x = 5 as dictated by the numbers to the left of 5 is 15, i.e.,

Similarly, when x approaches 5 from the right, f(x) should be taking value 15, i.e.,

© no N C tt E o R be T re pu

lim f ( x ) = 15 .

x →5+

Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are both equal to 15. Thus, lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 15 .

x → 5−

x →5

x →5

This conclusion about the limit being equal to 15 is somewhat strengthened by seeing the graph of this function which is given in Fig 2.16, Chapter 2. In this figure, we note that as x approaches 5 from either right or left, the graph of the function f(x) = x +10 approaches the point (5, 15).

We observe that the value of the function at x = 5 also happens to be equal to 15.

Illustration 2 Consider the function f(x) = x3. Let us try to find the limit of this function at x = 1. Proceeding as in the previous case, we tabulate the value of f(x) at x near 1. This is given in the Table 13.5.

Table 13.5

x

0.9

0.99

f(x)

0.729

0.970299

0.999

0.997002999

1.001

1.01

1.1

1.003003001

1.030301

1.331

From this table, we deduce that value of f(x) at x = 1 should be greater than 0.997002999 and less than 1.003003001 assuming nothing dramatic happens between

LIMITS AND DERIVATIVES

287

x = 0.999 and 1.001. It is reasonable to assume that the value of the f(x) at x = 1 as dictated by the numbers to the left of 1 is 1, i.e., lim f ( x ) = 1 .

x →1−

Similarly, when x approaches 1 from the right, f(x) should be taking value 1, i.e.,

he

lim f ( x ) = 1 .

x →1+

lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 1 .

x →1−

x →1

x →1

is

Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are both equal to 1. Thus,

© no N C tt E o R be T re pu

bl

This conclusion about the limit being equal to 1 is somewhat strengthened by seeing the graph of this function which is given in Fig 2.11, Chapter 2. In this figure, we note that as x approaches 1 from either right or left, the graph of the function f(x) = x3 approaches the point (1, 1).

We observe, again, that the value of the function at x = 1 also happens to be equal to 1.

Illustration 3 Consider the function f(x) = 3x. Let us try to find the limit of this function at x = 2. The following Table 13.6 is now self-explanatory.

Table 13.6

x

1.9

1.95

1.99

1.999

2.001

2.01

2.1

f(x)

5.7

5.85

5.97

5.997

6.003

6.03

6.3

As before we observe that as x approaches 2 from either left or right, the value of f(x) seem to approach 6. We record this as lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 6

x → 2−

x→2

x →2

Its graph shown in Fig 13.4 strengthens this

fact.

Here again we note that the value of the function at x = 2 coincides with the limit at x = 2.

Illustration 4 Consider the constant function f(x) = 3. Let us try to find its limit at x = 2. This function being the constant function takes the same

Fig 13.4

288

MATHEMATICS

value (3, in this case) everywhere, i.e., its value at points close to 2 is 3. Hence lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 3 x→ 2

x→2

x→2

Graph of f(x) = 3 is anyway the line parallel to x-axis passing through (0, 3) and is shown in Fig 2.9, Chapter 2. From this also it is clear that the required limit is 3. In

he

f ( x ) = 3 for any real number a. fact, it is easily observed that lim x→a

tabulate the values of f(x) near x = 1 in Table 13.7.

Table 13.7

0.99

0.999

f(x)

1.71

1.9701

1.997001

1.01

1.1

1.2

2.31

2.64

bl

0.9

2.0301

© no N C tt E o R be T re pu

x

is

f ( x ) . We Illustration 5 Consider the function f(x) = x2 + x. We want to find lim x →1

From this it is reasonable to deduce that lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 2 .

x →1−

x →1

x →1

From the graph of f(x) = x 2 + x shown in the Fig 13.5, it is clear that as x approaches 1, the graph approaches (1, 2).

Here, again we observe that the lim f (x) = f (1) x→1

Now, convince yourself of the following three facts:

Fig 13.5

lim x 2 = 1, lim x = 1 and lim x + 1 = 2 x →1

x →1

x →1

Then

lim x 2 + lim x = 1 + 1 = 2 = lim ⎡⎣ x 2 + x ⎤⎦ .

Also

lim x. lim ( x + 1) = 1.2 = 2 = lim ⎡⎣ x ( x + 1) ⎤⎦ = lim ⎡⎣ x 2 + x ⎤⎦ .

x →1

x →1

x →1

x →1

x →1

x →1

x →1

LIMITS AND DERIVATIVES

289

Illustration 6 Consider the function f(x) = sin x. We are interested in limπ sin x , x→

2

where the angle is measured in radians. π (Table 13.8). From 2

he

Here, we tabulate the (approximate) value of f(x) near this, we may deduce that π 2

x→

π 2

x→

π 2

.

is

lim− f ( x ) = lim+ f ( x ) = lim f ( x ) = 1

x→

Further, this is supported by the graph of f(x) = sin x which is given in the Fig 3.8 x→

π 2

© no N C tt E o R be T re pu

Table 13.8

bl

(Chapter 3). In this case too, we observe that lim sin x = 1.

x

π − 0.1 2

π − 0.01 2

π + 0.01 2

π + 0.1 2

f(x)

0.9950

0.9999

0.9999

0.9950

Illustration 7 Consider the function f(x) = x + cos x. We want to find the lim f (x). x→0

Here we tabulate the (approximate) value of f(x) near 0 (Table 13.9).

Table 13.9

x

– 0.1

– 0.01

– 0.001

0.001

0.01

0.1

f(x)

0.9850

0.98995

0.9989995

1.0009995

1.00995

1.0950

From the Table 13.9, we may deduce that

lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 1

x →0−

x →0

x →0

In this case too, we observe that lim f (x) = f (0) = 1. x→0

Now, can you convince yourself that

lim [ x + cos x ] = lim x + lim cos x is indeed true? x →0 x →0 x →0

290

MATHEMATICS

1 for x > 0 . We want to know lim f (x). x→0 x2 Here, observe that the domain of the function is given to be all positive real numbers. Hence, when we tabulate the values of f(x), it does not make sense to talk of x approaching 0 from the left. Below we tabulate the values of the function for positive x close to 0 (in this table n denotes any positive integer).

From the Table 13.10 given below, we see that as x tends to 0, f(x) becomes larger and larger. What we mean here is that the value of f(x) may be made larger than any given number.

Table 13.10

1

0.1

f(x)

1

100

10–n

10000

102n

© no N C tt E o R be T re pu

Mathematically, we say

0.01

bl

x

is

he

Illustration 8 Consider the function f ( x ) =

lim f ( x ) = + ∞ x →0

We also remark that we will not come across such limits in this course. f ( x ) , where Illustration 9 We want to find lim x →0

⎧ x − 2, x < 0 ⎪ f ( x) = ⎨ 0 , x = 0 ⎪ x + 2, x > 0 ⎩

As usual we make a table of x near 0 with f(x). Observe that for negative values of x we need to evaluate x – 2 and for positive values, we need to evaluate x + 2.

Table 13.11

x

– 0.1

– 0.01

– 0.001

0.001

0.01

0.1

f(x)

– 2.1

– 2.01

– 2.001

2.001

2.01

2.1

From the first three entries of the Table 13.11, we deduce that the value of the function is decreasing to –2 and hence. lim f ( x ) = −2

x →0−

LIMITS AND DERIVATIVES

291

From the last three entires of the table we deduce that the value of the function is increasing from 2 and hence lim f ( x ) = 2

x →0+

x ≠1 x =1

© no N C tt E o R be T re pu

⎧x + 2 f ( x) = ⎨ ⎩ 0

is

where

Fig 13.6

bl

f ( x) , Illustration 10 As a final illustration, we find lim x →1

he

Since the left and right hand limits at 0 do not coincide, we say that the limit of the function at 0 does not exist.

Graph of this function is given in the Fig13.6. Here, we remark that the value of the function at x = 0 is well defined and is, indeed, equal to 0, but the limit of the function at x = 0 is not even defined.

Table 13.12

x

0.9

0.99

0.999

1.001

1.01

1.1

f(x)

2.9

2.99

2.999

3.001

3.01

3.1

As usual we tabulate the values of f(x) for x near 1. From the values of f(x) for x less than 1, it seems that the function should take value 3 at x = 1., i.e., lim f ( x ) = 3 .

x →1−

Similarly, the value of f(x) should be 3 as dictated by values of f(x) at x greater than 1. i.e. lim f ( x ) = 3 .

x →1+

But then the left and right hand limits coincide and hence lim f ( x ) = lim+ f ( x ) = lim f ( x ) = 3 .

x →1−

x →1

x →1

Graph of function given in Fig 13.7 strengthens our deduction about the limit. Here, we

Fig 13.7

292

MATHEMATICS

note that in general, at a given point the value of the function and its limit may be different (even when both are defined).

he

13.3.1 Algebra of limits In the above illustrations, we have observed that the limiting process respects addition, subtraction, multiplication and division as long as the limits and functions under consideration are well defined. This is not a coincidence. In fact, below we formalise these as a theorem without proof.

Theorem 1 Let f and g be two functions such that both lim f (x) and lim g(x) exist. x →a

x →a

is

Then

(i) Limit of sum of two functions is sum of the limits of the functions, i.e., lim [f(x) + g (x)] = lim f(x) + lim g(x). x →a

x →a

bl

x →a

(ii) Limit of difference of two functions is difference of the limits of the functions, i.e., lim [f(x) – g(x)] = lim f(x) – lim g(x). x →a

x →a

© no N C tt E o R be T re pu

x →a

(iii) Limit of product of two functions is product of the limits of the functions, i.e., lim [f(x) . g(x)] = lim f(x). lim g(x). x →a

x →a

x →a

(iv) Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero), i.e.,

lim

x →a

$ Note

f ( x)

g ( x)

=

lim f ( x ) x →a

lim g ( x ) x→a

In particular as a special case of (iii), when g is the constant function

such that g(x) = λ , for some real number λ , we have

lim ⎡⎣( λ. f ) ( x ) ⎤⎦ = λ.lim f ( x ) . x →a x →a

In the next two subsections, we illustrate how to exploit this theorem to evaluate limits of special types of functions.

13.3.2 Limits of polynomials and rational functions A function f is said to be a polynomial function if f(x) is zero function or if f(x) = a0 + a1x + a2x2 +. . . + anxn, where ais are real numbers such that an ≠ 0 for some natural number n.

We know that lim x = a. Hence x →a

LIMITS AND DERIVATIVES

293

lim x 2 = lim ( x.x ) = lim x.lim x = a. a = a 2 x →a

x →a

x →a

x →a

An easy exercise in induction on n tells us that

lim x n = a n x →a

he

Now, let f ( x ) = a0 + a1 x + a2 x 2 + ... + an x n be a polynomial function. Thinking

lim f ( x ) x→ a

⎡ a0 + a1 x + a2 x 2 + ... + an x n ⎤ = lim ⎦ x →a ⎣

is

of each of a0 , a1 x, a2 x 2 ,..., an x n as a function, we have

bl

a0 + lim a1 x + lim a2 x 2 + ... + lim an x n = lim x →a x →a x →a x →a

© no N C tt E o R be T re pu

x + a2 lim x 2 + ... + an lim x n = a0 + a1 lim x →a x →a x →a = a0 + a1a + a2 a 2 + ... + an a n = f (a)

(Make sure that you understand the justification for each step in the above!)

g ( x) A function f is said to be a rational function, if f(x) = h x , where g(x) and h(x) ( )

are polynomials such that h(x) ≠ 0. Then

lim f ( x ) = lim x →a

x →a

g ( x) h( x)

=

lim g ( x ) x→a

lim h ( x )

=

x→a

g (a) h(a)

However, if h(a) = 0, there are two scenarios – (i) when g(a) ≠ 0 and (ii) when g(a) = 0. In the former case we say that the limit does not exist. In the latter case we can write g(x) = (x – a)k g1 (x), where k is the maximum of powers of (x – a) in g(x) Similarly, h(x) = (x – a)l h1 (x) as h (a) = 0. Now, if k > l, we have lim f ( x ) = x→a

lim g ( x ) x→a

lim h ( x ) x→a

lim ( x − a ) g1 ( x ) k

=

x→a

lim ( x − a ) h1 ( x ) l

x→a

294

MATHEMATICS

lim ( x − a )

=

x→a

( k −l )

g1 ( x )

lim h1 ( x )

=

x →a

0.g1 ( a ) h1 ( a )

=0

If k < l, the limit is not defined.

(iii)

⎡ x ( x + 1) ⎤⎦ (ii) lim x →3 ⎣

he

⎡ x3 − x 2 + 1⎤ Example 1 Find the limits: (i) lim ⎦ x →1 ⎣ lim ⎡1 + x + x 2 + ... + x10 ⎤⎦ .

x →−1 ⎣

© no N C tt E o R be T re pu

⎡ x ( x + 1) ⎤⎦ = 3 ( 3 + 1) = 3 ( 4 ) = 12 (ii) lim x →3 ⎣

bl

(i) lim [x3 – x2 + 1] = 13 – 12 + 1 = 1 x→1

is

Solution The required limits are all limits of some polynomial functions. Hence the limits are the values of the function at the prescribed points. We have

(iii)

lim ⎡1 + x + x 2 + ... + x10 ⎤⎦ = 1 + ( −1) + ( −1)2 + ... + ( −1)10

x →−1 ⎣

= 1 − 1 + 1... + 1 = 1 .

Example 2 Find the limits: ⎡ x2 + 1 ⎤ ⎢ ⎥ (i) lim x →1 x + 100 ⎣ ⎦

⎡ ⎤ x2 − 4 lim ⎥ (iii) x →2 ⎢ 3 2 ⎣ x − 4x + 4x ⎦

⎡ x3 − 4 x 2 + 4 x ⎤ ⎢ ⎥ (ii) lim x →2 x2 − 4 ⎣ ⎦

⎡ x3 − 2 x 2 ⎤ lim ⎥ (iv) x →2 ⎢ 2 ⎣ x − 5x + 6 ⎦

1 ⎡ x−2 ⎤ − 3 (v) lim .

2 2 ⎢ x →1 ⎣ x − x x − 3x + 2 x ⎥⎦

Solution All the functions under consideration are rational functions. Hence, we first

evaluate these functions at the prescribed points. If this is of the form

0 , we try to 0

rewrite the function cancelling the factors which are causing the limit to be of the form

0 .

0

LIMITS AND DERIVATIVES

295

x2 + 1 12 + 1 2 = = x →1 x + 100 1 + 100 101

(i) We have lim

Evaluating the function at 2, it is of the form

0 .

0

he

(ii)

x ( x − 2) x3 − 4 x 2 + 4 x lim Hence = lim 2 x → 2 ( x + 2 )( x − 2 ) x →2 x −4 2

is

as x ≠ 2

( x + 2)

2 ( 2 − 2) 2+2

=

0 =0.

4

© no N C tt E o R be T re pu

=

x ( x − 2)

bl

= lim x →2

(iii) Evaluating the function at 2, we get it of the form

Hence lim

x →2

0 .

0

( x + 2 )( x − 2 ) x2 − 4 lim 2 = x→2 x ( x − 2) x3 − 4 x 2 + 4 x

( x + 2) = 2 + 2 = 4 = lim x →2 x ( x − 2 ) 2 ( 2 − 2) 0

which is not defined.

(iv) Evaluating the function at 2, we get it of the form

Hence

0 .

0

x2 ( x − 2) x3 − 2 x 2 lim 2 = lim x → 2 ( x − 2 )( x − 3) x →2 x − 5 x + 6

( 2) 4 x2 = = = −4. ( x − 3) 2 − 3 −1 2

= lim x →2

296

MATHEMATICS

(v) First, we rewrite the function as a rational function.

(

)

⎤ ⎥ ⎥ ⎦

⎡ x−2 ⎤ 1 = ⎢ x x −1 − x x −1 x − 2 ⎥ ) ( )( ) ⎦⎥ ⎣⎢ (

bl

x2 − 4x + 3 = x x −1 x − 2 ( )( )

is

⎡ x2 − 4x + 4 − 1 ⎤ = ⎢ x x −1 x − 2 ⎥ )( ) ⎥⎦ ⎢⎣ (

0 .

0

© no N C tt E o R be T re pu Evaluating the function at 1, we get it of the form

Hence

he

⎡ 1 1 ⎡ x−2 ⎤ ⎢ x−2 − − = 2 ⎢ x 2 − x x3 − 3x 2 + 2 x ⎥ ⎢ x ( x − 1) x x − 3 x + 2 ⎣ ⎦ ⎣

⎡ x2 − 2 ⎤ x2 − 4x + 3 1 lim lim ⎢ 2 − 3 ⎥= x →1 x − x x − 3 x 2 + 2 x ⎦ x →1 x ( x − 1)( x − 2 ) ⎣

( x − 3)( x − 1) = lim x →1 x ( x − 1)( x − 2 )

x−3 1− 3 = lim = x →1 x ( x − 2 ) 1(1 − 2 ) = 2.

We remark that we could cancel the term (x – 1) in the above evaluation because x ≠1.

Evaluation of an important limit which will be used in the sequel is given as a theorem below.

Theorem 2 For any positive integer n, xn − an = na n −1 . x→a x − a

lim

Remark The expression in the above theorem for the limit is true even if n is any rational number and a is positive.

LIMITS AND DERIVATIVES

297

Proof Dividing (xn – an) by (x – a), we see that xn – an = (x–a) (xn–1 + xn–2 a + xn–3 a2 + ... + x an–2 + an–1) xn − an = lim (xn–1 + xn–2 a + xn–3 a2 + ... + x an–2 + an–1) x →a x − a x →a

Thus, lim

he

= an – l + a an–2 +. . . + an–2 (a) +an–l = an–1 + an – 1 +...+an–1 + an–1 (n terms)

Example 3 Evaluate:

x15 − 1 x →1 x10 − 1 Solution (i) We have (i) lim

1+ x −1 x

bl

(ii) lim

is

n −1 = na

© no N C tt E o R be T re pu

x →0

lim

x →1

15 10 x15 − 1 lim ⎡ x − 1 ÷ x − 1 ⎤ ⎢ ⎥ = x →1 x −1 ⎦ x10 − 1 ⎣ x −1

⎡ x15 − 1 ⎤ ⎡ x10 − 1 ⎤ lim lim ÷ ⎥ ⎢ ⎥ = x →1 ⎢ ⎣ x − 1 ⎦ x →1 ⎣ x − 1 ⎦

= 15 (1)14 ÷ 10(1)9 (by the theorem above) = 15 ÷ 10 =

3 2

(ii) Put y = 1 + x, so that y → 1 as x → 0.

Then

lim

x →0

y −1 1+ x −1 = lim y →1 y – 1 x 1

1

y 2 − 12 = lim y →1 y −1

=

1 1 1 2 −1 (1) (by the remark above) = 2 2

298

MATHEMATICS

13.4 Limits of Trigonometric Functions The following facts (stated as theorems) about functions in general come in handy in calculating limits of some trigonometric functions.

Theorem 3 Let f and g be two real valued functions with the same domain such that

he

f (x) ≤ g( x) for all x in the domain of definition, For some a, if both lim f(x) and x →a

© no N C tt E o R be T re pu

bl

is

lim g(x) exist, then lim f(x) ≤ lim g(x). This is illustrated in Fig 13.8. x →a x →a x →a

Fig 13.8

Theorem 4 (Sandwich Theorem) Let f, g and h be real functions such that f (x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a, if lim f(x) = l = lim h(x), then lim g(x) = l. This is illustrated in Fig 13.9. x→a x→a x→a

Fig 13.9

Given below is a beautiful geometric proof of the following important inequality relating trigonometric functions.

cos x <

sin x <1 x

for 0 < x <

π 2

(*)

LIMITS AND DERIVATIVES

299

Proof We know that sin (– x) = – sin x and cos( – x) = cos x. Hence, it is sufficient

π .

2 In the Fig 13.10, O is the centre of the unit circle such that

π . Line segments B A and 2 CD are perpendiculars to OA. Further, join AC. Then Area of ΔOAC < Area of sector OAC < Area of Δ OAB.

1 x 1 OA.CD < .π.(OA)2 < OA.AB .

2 2π 2 i.e., CD < x . OA < AB. From Δ OCD,

Fig 13.10

is

the angle AOC is x radians and 0 < x <

he

to prove the inequality for 0 < x <

bl

i.e.,

CD AB (since OC = OA) and hence CD = OA sin x. Also tan x = and OA OA hence AB = OA. tan x. Thus OA sin x < OA. x < OA. tan x.

Since length OA is positive, we have sin x < x < tan x.

© no N C tt E o R be T re pu

sin x =

Since 0 < x < 1<

π , sinx is positive and thus by dividing throughout by sin x, we have 2

x 1 < . Taking reciprocals throughout, we have sin x cos x

sin x <1 x which complete the proof. cos x <

Theorem 5 The following are two important limits.

sin x =1. x →0 x

(i) lim

1 − cos x =0. x →0 x

(ii) lim

sin x is sandwiched between the x function cos x and the constant function which takes value 1.

Proof (i) The inequality in (*) says that the function

300

MATHEMATICS

Further, since lim cos x = 1, we see that the proof of (i) of the theorem is x→0 complete by sandwich theorem.

Then

is

⎛x⎞ ⎛x⎞ 2sin 2 ⎜ ⎟ sin ⎜ ⎟ ⎝ 2 ⎠ = lim ⎝ 2 ⎠ .sin ⎛ x ⎞ ⎜2⎟ x →0 x x ⎝ ⎠ 2

bl

1 − cos x lim = lim x →0 x →0 x

he

⎛x⎞ To prove (ii), we recall the trigonometric identity 1 – cos x = 2 sin2 ⎜ ⎟ . ⎝2⎠

© no N C tt E o R be T re pu

⎛x⎞ sin ⎜ ⎟ 2 ⎛x⎞ = lim ⎝ ⎠ .lim sin ⎜ ⎟ = 1.0 = 0 x →0 x →0 x ⎝2⎠ 2

Observe that we have implicitly used the fact that x → 0 is equivalent to

may be justified by putting y =

Example 4 Evaluate:

Solution (i)

lim

x →0

x .

2

(i) lim

sin 4 x

x →0 sin 2 x

(ii) lim

x →0

tan x x

⎡ sin 4 x 2 x ⎤ sin 4 x = lim ⎢ . .2⎥ x → 0 sin 2 x ⎣ 4 x sin 2 x ⎦

⎡ sin 4 x ⎤ ⎡ sin 2 x ⎤ = 2.lim ⎥÷⎢ ⎥ x →0 ⎢ ⎣ 4x ⎦ ⎣ 2x ⎦

⎡ sin 4 x ⎤ ⎡ sin 2 x ⎤ ÷ lim ⎢ = 2. 4lim ⎥ ⎥ x →0 ⎢ 2 x → 0 ⎣ 4x ⎦ ⎣ 2x ⎦

= 2.1.1 = 2 (as x → 0, 4x → 0 and 2x → 0)

x → 0 . This 2

LIMITS AND DERIVATIVES

(ii) We have lim x →0

301

sin x sin x 1 tan x . lim = lim = lim = 1.1 = 1 x → 0 x → 0 x → 0 cos x x cos x x x

A general rule that needs to be kept in mind while evaluating limits is the following.

he

f ( x) exists and we want to evaluate this. First we check Say, given that the limit lim x→a g ( x )

g ( x)

q ( x ) , where q(x) ≠ 0.

f ( x)

=

p (a)

© no N C tt E o R be T re pu

lim

p ( x)

=

bl

f ( x)

is

the value of f (a) and g(a). If both are 0, then we see if we can get the factor which is causing the terms to vanish, i.e., see if we can write f(x) = f1 (x) f2(x) so that f1 (a) = 0 and f2 (a) ≠ 0. Similarly, we write g(x) = g1 (x) g2(x), where g1(a) = 0 and g2(a) ≠ 0. Cancel out the common factors from f(x) and g(x) (if possible) and write

Then

x →a

g ( x)

q(a) .

EXERCISE 13.1

Evaluate the following limits in Exercises 1 to 22. x+3 1. lim x →3

4x + 3 x →4 x − 2

4. lim

7. lim

x →2

3x 2 − x − 10 x2 − 4

22 ⎞ ⎛ x− ⎟ 2. lim ⎜ x→π ⎝ 7 ⎠

πr 2 3. lim r →1

( x + 1)5 − 1

5. lim

x10 + x5 + 1 x −1

6. lim

x→ 0

x

8. lim

x 4 − 81 2 x2 − 5x − 3

9. lim

ax + b cx + 1

x → −1

x →3

x →0

1

10. lim z →1

z 3 −1 1

z 6 −1

1 1 + 12. lim x 2 x →−2 x + 2

ax 2 + bx + c ,a+b+c ≠0 x →1 cx 2 + bx + a

11. lim

13. lim

x →0

sin ax bx

14. lim

x →0

sin ax , a, b ≠ 0 sin bx

MATHEMATICS

sin ( π − x ) 15. lim x→π π ( π − x )

16. lim

x →0

cos 2 x − 1 x →0 cos x − 1

cos x π−x

17. lim

18. lim

ax + x cos x b sin x

x sec x 19. lim x →0

20. lim

sin ax + bx a, b, a + b ≠ 0 , ax + sin bx

(cosec x − cot x) 21. lim x →0

x →0

lim

tan 2 x π x− 2

bl

22.

π x→ 2

is

x →0

he

302

© no N C tt E o R be T re pu

⎧ 2 x + 3, x ≤ 0 f ( x ) and lim f ( x ) , where f ( x ) = ⎨ 23. Find lim x →0 x →1 ⎩3 ( x + 1) , x > 0

⎧⎪ x 2 − 1, x ≤ 1 = f x lim f x ( ) ⎨ 2 24. Find x →1 ( ) , where ⎪⎩ − x − 1, x > 1

⎧| x | ⎪ , x≠0 f ( x ) , where f ( x ) = ⎨ x 25. Evaluate lim x →0 ⎪⎩ 0, x = 0

⎧ x ⎪ , x≠0 f ( x ) , where f ( x ) = ⎨ | x | 26. Find lim x →0 ⎪ 0, x = 0 ⎩ f ( x ) , where f ( x ) = | x | −5 27. Find lim x →5 ⎧ a + bx , ⎪ 28. Suppose f ( x ) = ⎨4, ⎪ b − ax , ⎩

x <1 x =1 x >1

and if lim f (x) = f (1) what are possible values of a and b? x→1

LIMITS AND DERIVATIVES

303

29. Let a1, a2, . . ., an be fixed real numbers and define a function f ( x ) = ( x − a1 ) ( x − a2 ) ... ( x − an ) .

What is xlim f (x) ? For some a ≠ a1, a2, ..., an, compute lim f (x). x→a →a

he

1

31. If the function f(x) satisfies lim

x2 − 1

= π , evaluate lim f ( x ) . x →1

© no N C tt E o R be T re pu

x →1

f ( x) − 2

bl

For what value (s) of a does lim f (x) exists? x →a

is

⎧ x + 1, x < 0 ⎪ x=0 .

30. If f ( x ) = ⎨ 0, ⎪ x − 1, x > 0 ⎩

32.

⎧ mx 2 + n , x<0 ⎪ f ( x) If f ( x ) = ⎨ nx + m , 0 ≤ x ≤ 1 . For what integers m and n does both lim x →0 ⎪ 3 x >1 ⎩ nx + m ,

f ( x ) exist? and lim x →1

13.5 Derivatives

We have seen in the Section 13.2, that by knowing the position of a body at various time intervals it is possible to find the rate at which the position of the body is changing.

It is of very general interest to know a certain parameter at various instants of time and try to finding the rate at which it is changing. There are several real life situations where such a process needs to be carried out. For instance, people maintaining a reservoir need to know when will a reservoir overflow knowing the depth of the water at several instances of time, Rocket Scientists need to compute the precise velocity with which the satellite needs to be shot out from the rocket knowing the height of the rocket at various times. Financial institutions need to predict the changes in the value of a particular stock knowing its present value. In these, and many such cases it is desirable to know how a particular parameter is changing with respect to some other parameter.

The heart of the matter is derivative of a function at a given point in its domain of definition.

304

MATHEMATICS

Definition 1 Suppose f is a real valued function and a is a point in its domain of definition. The derivative of f at a is defined by lim

f (a + h) − f (a)

h →0

h provided this limit exists. Derivative of f (x) at a is denoted by f′(a).

Example 5 Find the derivative at x = 2 of the function f(x) = 3x.

f′ (2) = lim

f ( 2 + h ) − f ( 2)

h →0

3( 2 + h ) − 3( 2) h

6 + 3h − 6 3h = lim = lim 3 = 3 . h → 0 h h h →0

© no N C tt E o R be T re pu

h →0

h →0

bl

= lim

h

= lim

is

Solution We have

he

Observe that f′ (a) quantifies the change in f(x) at a with respect to x.

The derivative of the function 3x at x = 2 is 3.

Example 6 Find the derivative of the function f(x) = 2x2 + 3x – 5 at x = –1. Also prove that f ′ (0) + 3f ′ ( –1) = 0.

Solution We first find the derivatives of f(x) at x = –1 and at x = 0. We have f ' ( −1) = lim

h →0

f ( −1 + h ) − f ( −1) h

⎡ 2 ( −1 + h )2 + 3 ( −1 + h ) − 5⎤ − ⎡ 2 ( −1)2 + 3 ( −1) − 5⎤ ⎦ ⎣ ⎦ = lim ⎣ h →0 h = lim

h →0

and

f ' ( 0 ) = lim

h →0

2h 2 − h = lim ( 2h − 1) = 2 ( 0 ) − 1 = −1 h →0 h f (0 + h ) − f (0) h

⎡ 2 ( 0 + h )2 + 3 ( 0 + h ) − 5⎤ − ⎡ 2 ( 0 )2 + 3 ( 0 ) − 5⎤ ⎦ ⎣ ⎦ = lim ⎣ h →0 h

LIMITS AND DERIVATIVES

305

2h 2 + 3h = lim ( 2h + 3) = 2 ( 0 ) + 3 = 3 h →0 h →0 h

= lim Clearly

f ' ( 0 ) + 3 f ' ( −1) = 0

he

Remark At this stage note that evaluating derivative at a point involves effective use of various rules, limits are subjected to. The following illustrates this.

Example 7 Find the derivative of sin x at x = 0. f ( 0 + h ) − f ( 0)

f ′(0) = lim

h →0

h sin ( 0 + h ) − sin ( 0 )

h →0

sin h =1 h →0 h

= lim

© no N C tt E o R be T re pu

h

bl

= lim

is

Solution Let f(x) = sin x. Then

Example 8 Find the derivative of f(x) = 3 at x = 0 and at x = 3.

Solution Since the derivative measures the change in function, intuitively it is clear that the derivative of the constant function must be zero at every point. This is indeed, supported by the following computation. f ' ( 0 ) = lim

f (0 + h) − f (0)

h →0

Similarly

f ' ( 3) = lim

h→0

h

f ( 3 + h ) − f ( 3) h

= lim

3−3 0 = lim = 0 . h →0 h h

= lim

3−3 =0. h

h →0

h →0

We now present a geometric interpretation of derivative of a function at a point. Let y = f(x) be a function and let P = (a, f(a)) and Q = (a + h, f(a + h) be two points close to each other on the graph of this function. The Fig 13.11 is now self explanatory.

Fig 13.11

306

MATHEMATICS

We know that f ′ ( a ) = lim

f (a + h) − f (a)

h →0

f (a + h) − f (a)

he

h From the triangle PQR, it is clear that the ratio whose limit we are taking is precisely equal to tan(QPR) which is the slope of the chord PQ. In the limiting process, as h tends to 0, the point Q tends to P and we have

QR h→0 Q → P PR h This is equivalent to the fact that the chord PQ tends to the tangent at P of the curve y = f(x). Thus the limit turns out to be equal to the slope of the tangent. Hence = lim

is

lim

f ′ ( a ) = tan ψ .

© no N C tt E o R be T re pu

bl

For a given function f we can find the derivative at every point. If the derivative exists at every point, it defines a new function called the derivative of f . Formally, we define derivative of a function as follows.

Definition 2 Suppose f is a real valued function, the function defined by lim

f ( x + h) − f ( x)

h →0

h wherever the limit exists is defined to be the derivative of f at x and is denoted by f′(x). This definition of derivative is also called the first principle of derivative.

Thus

f ' ( x ) = lim

f ( x + h) − f ( x)

h →0

h Clearly the domain of definition of f′ (x) is wherever the above limit exists. There are different notations for derivative of a function. Sometimes f′ (x) is denoted by

d dy f ( x ) ) or if y = f(x), it is denoted by ( . This is referred to as derivative of f(x) dx dx or y with respect to x. It is also denoted by D (f (x) ). Further, derivative of f at x = a is also denoted by

d f ( x) dx

a

or

df dx

a

⎛ df ⎞ or even ⎜ ⎟ . ⎝ dx ⎠ x = a

Example 9 Find the derivative of f(x) = 10x.

Solution Since f′ ( x) = lim

h →0

f ( x + h) − f ( x) h

LIMITS AND DERIVATIVES

= lim

307

10 ( x + h ) − 10 ( x )

h →0

h

10h (10 ) = 10 . = lim h→ 0 h →0 h

he

= lim

Example 10 Find the derivative of f(x) = x2. f ( x + h) − f ( x)

h →0

( x + h )2 − ( x )2

( h + 2x ) = 2x = lim h →0

h

bl

= lim h →0

h

is

Solution We have, f ′(x) = lim

© no N C tt E o R be T re pu

Example 11 Find the derivative of the constant function f (x) = a for a fixed real number a.

Solution We have, f ′(x) = lim

f ( x + h) − f ( x)

h →0

= lim

h →0

h

a−a 0 = lim = 0 as h ≠ 0 h →0 h h

Example 12 Find the derivative of f(x) = Solution We have f ′(x) = lim

1 x

f ( x + h) − f ( x)

h →0

h

1 1 – = lim ( x + h) x h →0 h

1 ⎡ x − ( x + h) ⎤ ⎢ ⎥ = lim h →0 h ⎢⎣ x ( x + h ) ⎦⎥

−1 1 ⎡ −h ⎤ ⎢ ⎥ = lim = lim h → 0 h →0 h x ( x + h ) x ( x + h) ⎢⎣ ⎥⎦

= −

1 x2

308

MATHEMATICS

13.5.1 Algebra of derivative of functions Since the very definition of derivatives involve limits in a rather direct fashion, we expect the rules for derivatives to follow closely that of limits. We collect these in the following theorem.

Theorem 5 Let f and g be two functions such that their derivatives are defined in a common domain. Then

d d d ⎡⎣ f ( x) + g ( x) ⎤⎦ = f ( x) + g( x) . dx dx dx

Derivative of difference of two functions is difference of the derivatives of the functions.

bl

(ii)

he

Derivative of sum of two functions is sum of the derivatives of the functions.

is

(i)

© no N C tt E o R be T re pu

d d d ⎡⎣ f ( x ) − g ( x ) ⎤⎦ = f ( x) − g ( x) . dx dx dx (iii)

Derivative of product of two functions is given by the following product rule.

d d d ⎡⎣ f ( x ) . g ( x ) ⎤⎦ = f ( x) . g ( x) + f ( x) . g ( x) dx dx dx

(iv)

Derivative of quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero). d d f ( x) . g ( x) − f ( x) g ( x) d ⎛ f ( x) ⎞ dx dx ⎜ ⎟= dx ⎝ g ( x) ⎠ ( g ( x) )2

The proofs of these follow essentially from the analogous theorem for limits. We will not prove these here. As in the case of limits this theorem tells us how to compute derivatives of special types of functions. The last two statements in the theorem may be restated in the following fashion which aids in recalling them easily: Let u = f ( x ) and v = g (x). Then

( uv )′

= u′v + uv′

This is referred to a Leibnitz rule for differentiating product of functions or the product rule. Similarly, the quotient rule is

LIMITS AND DERIVATIVES

309

⎛ u ⎞′ u ′v − uv′ ⎜ ⎟ = v2 ⎝v⎠

Now, let us tackle derivatives of some standard functions.

It is easy to see that the derivative of the function f(x) = x is the constant f ( x + h) − f ( x)

h →0

h

x+h−x h →0 h

he

function 1. This is because f ′ ( x ) = lim

= lim

is

=1 . = lim1 h→0

df ( x) d = dx dx

( x + ... + x )

bl

We use this and the above theorem to compute the derivative of f(x) = 10x = x + .... + x (ten terms). By (i) of the above theorem (ten terms)

© no N C tt E o R be T re pu

d d x +...+ x (ten terms) dx dx = 1 + ... + 1 (ten terms) = 10.

We note that this limit may be evaluated using product rule too. Write f(x) = 10x = uv, where u is the constant function taking value 10 everywhere and v(x) = x. Here, f(x) = 10x = uv we know that the derivative of u equals 0. Also derivative of v(x) = x equals 1. Thus by the product rule we have =

f ′ ( x ) = (10 x )′ = ( uv )′ = u ′v + uv′ = 0.x + 10.1 = 10

On similar lines the derivative of f(x) = x 2 may be evaluated. We have f(x) = x2 = x .x and hence

df d d d ( x.x ) = ( x ) .x + x. ( x ) = dx dx dx dx = 1.x + x.1 = 2 x .

More generally, we have the following theorem.

Theorem 6 Derivative of f(x) = xn is nxn – 1 for any positive integer n.

Proof By definition of the derivative function, we have f ' ( x ) = lim

h →0

f ( x + h) − f ( x) h

= lim

h→0

( x + h )n − x n h

.

310

MATHEMATICS

Binomial theorem tells that (x + h)n = hence (x + h) – x = h(nx n

n–1

n

+... + h

(

n

)

C0 x n +

( C )x n

1

n −1

h + ... +

(

n

)

C n h n and

). Thus

n–1

df ( x) ( x + h ) − xn = lim dx h →0 h

(

h →0

)

h

(

)

nx n −1 + ... + h n −1 = nx n −1 . = lim h →0

is

= lim

h nx n −1 + .... + h n −1

he

n

( )

(

)

© no N C tt E o R be T re pu

d n d x.x n −1 x = dx dx

bl

Alternatively, we may also prove this by induction on n and the product rule as follows. The result is true for n = 1, which has been proved earlier. We have

=

(

)

(

)

d d ( x ) . x n−1 + x. x n−1 (by product rule) dx dx

= 1.x

n −1

(

)

+ x. ( n − 1) x n − 2 (by induction hypothesis)

= x n −1 + ( n − 1) x n −1 = nx n −1 .

Remark The above theorem is true for all powers of x, i.e., n can be any real number (but we will not prove it here).

13.5.2 Derivative of polynomials and trigonometric functions We start with the

following theorem which tells us the derivative of a polynomial function.

Theorem 7 Let f(x) = an x n + an −1 x n −1 + .... + a1 x + a0 be a polynomial function, where

ai s are all real numbers and an ≠ 0. Then, the derivative function is given by

df ( x) = nan x n −1 + ( n − 1) an −1 x x − 2 + ... + 2a2 x + a1 . dx Proof of this theorem is just putting together part (i) of Theorem 5 and Theorem 6.

Example 13 Compute the derivative of 6x100 – x55 + x.

Solution A direct application of the above theorem tells that the derivative of the

above function is 600 x 99 − 55 x 54 + 1 .

LIMITS AND DERIVATIVES

311

Example 14 Find the derivative of f(x) = 1 + x + x2 + x3 +... + x50 at x = 1.

Solution A direct application of the above Theorem 6 tells that the derivative of the above function is 1 + 2x + 3x2 + . . . + 50x49. At x = 1 the value of this function equals

( 50 )( 51) 2

= 1275.

he

1 + 2(1) + 3(1)2 + .. . + 50(1)49 = 1 + 2 + 3 + . . . + 50 =

x +1 x Solution Clearly this function is defined everywhere except at x = 0. We use the quotient rule with u = x + 1 and v = x. Hence u′ = 1 and v′ = 1. Therefore

is

Example 15 Find the derivative of f(x) =

© no N C tt E o R be T re pu

Example 16 Compute the derivative of sin x.

bl

df ( x) d ⎛ x + 1 ⎞ d ⎛ u ⎞ u ′v − uv′ 1( x ) − ( x + 1)1 1 = ⎜ = =− 2 ⎟= ⎜ ⎟ = 2 2 dx dx ⎝ x ⎠ dx ⎝ v ⎠ v x x

Solution Let f(x) = sin x. Then

f ( x + h) − f ( x) sin ( x + h ) − sin ( x ) df ( x) = lim = lim h →0 h →0 dx h h

⎛ 2x + h ⎞ ⎛ h ⎞ 2cos ⎜ ⎟ sin ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ (using formula for sin A – sin B) = lim h →0 h h sin h⎞ ⎛ 2 = cos x.1 = cos x lim cos ⎜ x + ⎟ .lim . = h →0 h → 0 h 2 ⎝ ⎠ 2

Example 17 Compute the derivative of tan x.

Solution Let f(x) = tan x. Then

f ( x + h) − f ( x) tan ( x + h ) − tan ( x ) df ( x) = lim = lim h →0 h →0 dx h h

1 ⎡ sin ( x + h ) sin x ⎤ lim = h→0 h ⎢ cos x + h − cos x ⎥ ( ) ⎢⎣ ⎥⎦

312

MATHEMATICS

⎡ sin ( x + h ) cos x − cos ( x + h ) sin x ⎤ ⎢ ⎥ = lim h →0 h cos ( x + h ) cos x ⎢⎣ ⎥⎦

1 = sec 2 x . cos 2 x

bl

= 1.

sin h 1 .lim h h→0 cos ( x + h ) cos x

is

= lim h →0

he

sin ( x + h − x ) = lim (using formula for sin (A + B)) h →0 h cos ( x + h ) cos x

Example 18 Compute the derivative of f(x) = sin2 x.

© no N C tt E o R be T re pu

Solution We use the Leibnitz product rule to evaluate this.

df ( x) d = ( sin x sin x ) dx dx

= ( sin x )′ sin x + sin x ( sin x )′

= ( cos x ) sin x + sin x ( cos x )

= 2sin x cos x = sin 2 x .

EXERCISE 13.2

1.

2.

3.

4.

Find the derivative of x2 – 2 at x = 10.

Find the derivative of 99x at x = l00.

Find the derivative of x at x = 1.

Find the derivative of the following functions from first principle. (i)

x 3 − 27

1 x2 5. For the function

(iii)

f ( x) =

(ii)

( x − 1)( x − 2 )

(iv)

x +1 x −1

x100 x99 x2 + +. . .+ + x +1 .

100 99 2

LIMITS AND DERIVATIVES

313

Prove that f ′ (1) = 100 f ′ ( 0 ) .

6. Find the derivative of x n + ax n −1 + a 2 x n − 2 + . . . + a n −1 x + a n for some fixed real number a.

7. For some constants a and b, find the derivative of (ii)

8. Find the derivative of

( ax

2

+b

)

2

(iii)

xn − an for some constant a. x−a

9. Find the derivative of

3 4

(ii)

x −3 ( 5 + 3 x )

(v)

x −4 3 − 4 x −5

3

) ( x − 1)

+ 3x − 1

(

5 −9 (iv) x 3 − 6 x

)

© no N C tt E o R be T re pu

(iii)

(5x

bl

(i) 2 x −

(

)

(vi)

x−a x−b

he

( x − a) ( x − b)

is

(i)

2 x2 − x + 1 3x − 1

10. Find the derivative of cos x from first principle.

11. Find the derivative of the following functions: (i) sin x cos x (iv) cosec x

(ii) sec x

(iii) 5sec x + 4 cos x

(v) 3cot x + 5cosec x

(vi) 5sin x − 6 cos x + 7

(vii) 2 tan x − 7 sec x

Miscellaneous Examples

Example 19 Find the derivative of f from the first principle, where f is given by (i) f (x) =

2x + 3 x−2

(ii)

f (x) = x +

1 x

Solution (i) Note that function is not defined at x = 2. But, we have

f ′ ( x ) = lim

h →0

f ( x + h) − f ( x) h

2( x + h) + 3 2x + 3 − x−2 = lim x + h − 2 h →0 h

MATHEMATICS

= lim h →0

( 2 x + 2h + 3)( x − 2 ) − ( 2 x + 3)( x + h − 2 ) h ( x − 2 )( x + h − 2 )

= lim h →0

( 2 x + 3)( x − 2 ) + 2h ( x − 2 ) − ( 2 x + 3)( x − 2 ) − h ( 2 x + 3) h ( x − 2 )( x + h − 2 )

–7 7 =− = lim h →0 ( x − 2 ) ( x + h − 2 ) ( x − 2 )2

is

1 ⎞ ⎛ 1⎞ ⎛ x+h+ −⎜x+ ⎟ ⎜ ⎟ f ( x + h) − f ( x) x+h⎠ ⎝ x⎠ = lim ⎝ = lim h →0 h →0 h h

© no N C tt E o R be T re pu

f ′( x)

bl

(ii)

Again, note that the function f ′ is also not defined at x = 2.

The function is not defined at x = 0. But, we have

he

314

= lim h →0

1⎡ 1 1⎤ h+ − ⎥ ⎢ h⎣ x + h x⎦

⎞⎤ 1⎡ x − x −h⎤ 1⎡ ⎛ 1 lim h lim h 1 + = − ⎢ ⎥ ⎜ ⎟ ⎢ ⎥ = h →0 h x ( x + h ) ⎦⎥ h→0 h ⎣⎢ ⎜⎝ x ( x + h ) ⎠⎟ ⎦⎥ ⎣⎢ ⎡ ⎤ 1 1 ⎢1 − ⎥ =1− 2 = lim h →0 x ⎣⎢ x ( x + h ) ⎦⎥

Again, note that the function f ′ is not defined at x = 0.

Example 20 Find the derivative of f(x) from the first principle, where f(x) is (ii) x sin x (i) sin x + cos x Solution (i) we have f ' ( x ) = = lim

h →0

= lim

h →0

f ( x + h) − f ( x) h

sin ( x + h ) + cos ( x + h ) − sin x − cos x h

sin x cos h + cos x sin h + cos x cos h − sin x sin h − sin x − cos x h

LIMITS AND DERIVATIVES

sin h ( cos x − sin x ) + sin x ( cos h − 1) + cos x ( cos h − 1)

= lim

h→0

h

h →0

( cos h − 1) sin h ( cos h − 1) sin x ( cos x − sin x ) + lim + lim cos x h → 0 h →0 h h h

= cos x − sin x h→0

= lim

( x + h ) sin ( x + h ) − x sin x

h →0

h

h

( x + h )( sin x cos h + sin h cos x ) − x sin x

h →0

= lim

= lim

h

is

f ( x + h) − f ( x)

x sin x ( cos h − 1) + x cos x sin h + h ( sin x cos h + sin h cos x )

h →0

h

© no N C tt E o R be T re pu

x sin x ( cos h − 1)

bl

f ' ( x ) = lim

= lim

he

= lim

(ii)

315

h →0

h

+ lim h→0 x cos x

sin h + lim ( sin x cos h + sin h cos x ) h →0 h

= x cos x + sin x Example 21 Compute derivative of

(i) f(x) = sin 2x

(ii) g(x) = cot x

Solution (i) Recall the trigonometric formula sin 2x = 2 sin x cos x. Thus

df ( x) d d ( 2sin x cos x ) = 2 ( sin x cos x ) = dx dx dx = 2 ⎡( sin x )′ cos x + sin x ( cos x )′ ⎤ ⎣⎢ ⎦⎥

= 2 ⎡⎣( cos x ) cos x + sin x ( − sin x ) ⎤⎦

(

= 2 cos 2 x − sin 2 x

(ii) By definition, g(x) = cot x =

wherever it is defined.

)

cos x . We use the quotient rule on this function sin x

dg d d ⎛ cos x ⎞ = (cot x) = ⎜ dx dx dx ⎝ sin x ⎟⎠

MATHEMATICS

=

(cos x)′ (sin x) − (cos x) (sin x)′ (sin x) 2

=

(− sin x)(sin x) − (cos x)(cos x) (sin x)2

= −

he

316

sin 2 x + cos 2 x = − cosec 2 x 2 sin x

bl

© no N C tt E o R be T re pu

d d ⎛ 1 ⎞ dg (cot x) = = dx dx ⎜⎝ tan x ⎟⎠ dx

is

1 . Here, we use the fact tan x that the derivative of tan x is sec2 x which we saw in Example 17 and also that the derivative of the constant function is 0.

Alternatively, this may be computed by noting that cot x =

=

(1)′ (tan x) − (1)(tan x)′ (tan x) 2

(0) (tan x) − (sec x) 2 = (tan x) 2

=

− sec 2 x = − cosec 2 x tan 2 x

Example 22 Find the derivative of (i)

x5 − cos x sin x

Solution (i) Let h( x) =

(ii)

x + cos x tan x

x5 − cos x . We use the quotient rule on this function wherever sin x

it is defined.

h′( x) =

( x5 − cos x)′ sin x − ( x 5 − cos x) (sin x)′ (sin x) 2

LIMITS AND DERIVATIVES

=

317

(5 x 4 + sin x)sin x − ( x5 − cos x)cos x sin 2 x

h′( x) =

x + cos x wherever it is defined. tan x

( x + cos x)′ tan x − ( x + cos x)(tan x)′ (tan x)2

bl

(1 − sin x) tan x − ( x + cos x)sec 2 x = (tan x) 2

is

(ii) We use quotient rule on the function

he

− x 5 cos x + 5 x 4 sin x + 1 = (sin x) 2

© no N C tt E o R be T re pu

Miscellaneous Exercise on Chapter 13

1. Find the derivative of the following functions from first principle:

π ) 8 Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (i) − x

(ii) (− x) −1

(iii) sin (x + 1)

(iv) cos (x –

2. (x + a)

⎛r ⎞ 3. (px + q) ⎜ + s ⎟ ⎝x ⎠

4. ( ax + b )( cx + d )

ax + b 5. cx + d

1 x 6.

1 1− x

7.

8.

ax + b px + qx + r 2

1+

9.

px 2 + qx + r ax + b

10.

2

1 ax + bx + c 2

a b − 2 + cos x 4 x x

11. 4 x − 2

12. (ax + b) n

13. (ax + b) n (cx + d ) m

14. sin (x + a)

15. cosec x cot x

16.

cos x 1 + sin x

318

MATHEMATICS

17.

sin x + cos x sin x − cos x

18.

sec x − 1 sec x + 1

19. sin n x

20.

a + b sin x c + d cos x

21.

sin( x + a) cos x

22. x 4 (5sin x − 3cos x)

23.

(x

24.

( ax

+ sin x

25.

( x + cos x ) ( x − tan x )

4 x + 5sin x 26.

3x + 7cos x

28.

x 1 + tan x

29.

) ( p + q cos x )

⎛π ⎞ x 2 cos ⎜ ⎟ ⎝4⎠ 27. sin x

( x + sec x ) ( x − tan x )

30.

x sin n x

© no N C tt E o R be T re pu

Summary

he

2

is

)

+ 1 cos x

bl

2

The expected value of the function as dictated by the points to the left of a point defines the left hand limit of the function at that point. Similarly the right hand limit.

Limit of a function at a point is the common value of the left and right hand limits, if they coincide.

For a function f and a real number a,

lim f(x) and f (a) may not be same (In x →a

fact, one may be defined and not the other one).

For functions f and g the following holds: lim [ f ( x) ± g ( x) ] = lim f ( x) ± lim g ( x) x →a

x→a

x →a

lim [ f ( x). g ( x) ] = lim f ( x ).lim g ( x ) x→a

x→a

x→a

f ( x) ⎡ f ( x ) ⎤ lim x→a lim ⎢ = ⎥ x→a g ( x) g ( x) ⎣ ⎦ lim x →a

Following are some of the standard limits xn − an = na n −1 x →a x − a

lim

LIMITS AND DERIVATIVES

lim

x →0

319

sin x =1 x

1 − cos x =0 x →0 x The derivative of a function f at a is defined by f ( a + h) − f ( a ) h Derivative of a function f at any point x is defined by

he

lim

f ′(a ) = lim

(u ± v )′ = u ′ ± v′

© no N C tt E o R be T re pu

(uv )′ = u ′v + uv′

bl

df ( x) f ( x + h) − f ( x ) = lim h → 0 dx h For functions u and v the following holds: f ′( x) =

is

h →0

⎛ u ⎞′ u ′v − uv′ provided all are defined. ⎜ ⎟ = v2 ⎝v⎠

Following are some of the standard derivatives. d n ( x ) = nx n −1 dx

d (sin x) = cos x dx

d (cos x) = − sin x dx

Historical Note

In the history of mathematics two names are prominent to share the credit for inventing calculus, Issac Newton (1642 – 1727) and G.W. Leibnitz (1646 – 1717).

Both of them independently invented calculus around the seventeenth century.

After the advent of calculus many mathematicians contributed for further development of calculus. The rigorous concept is mainly attributed to the great

320

MATHEMATICS

mathematicians, A.L. Cauchy, J.L.Lagrange and Karl Weierstrass. Cauchy gave the foundation of calculus as we have now generally accepted in our textbooks.

Cauchy used D’ Alembert’s limit concept to define the derivative of a function.

Starting with definition of a limit, Cauchy gave examples such as the limit of

α

Δ y f ( x + i ) − f ( x) , and called the limit for for α = 0. He wrote Δ x = i

he

sin α

i → 0, the “function derive’e, y′ for f ′ (x)”.

bl

is

Before 1900, it was thought that calculus is quite difficult to teach. So calculus became beyond the reach of youngsters. But just in 1900, John Perry and others in England started propagating the view that essential ideas and methods of calculus were simple and could be taught even in schools. F.L. Griffin, pioneered the teaching of calculus to first year students. This was regarded as one of the most daring act in those days.

© no N C tt E o R be T re pu

Today not only the mathematics but many other subjects such as Physics, Chemistry, Economics and Biological Sciences are enjoying the fruits of calculus.

— —